Question

A solution is made by dissolving 25.5 grams of glucose (C6H12O6) in 398 grams of water. What is the freezing point depression of the solvent if the freezing point constant is -1.86 °C/m?

Answer 1

[delta]T=i Kf m?

Answer 2

Gotta use the freezing point equation: \(\sf \Delta T=i*m*K_f\)

where \(\Delta T\) is the change in temp, \(i\) is the van't hoff constant, m is the molality, and Kf is the depression constant (given in the question).

(we can ignore \(i\) because glucose doesnt dissociate into more particles in water).

First find the molality though

Answer 3

yep

Answer 4

M=number of moles/Volume

Answer 5

that's molaRity, this is molaLity

Answer 6

oh my b

Answer 7

\(\sf molality=\dfrac{moles~of~solute}{kg~of~solution}\)

Answer 8

lol no worries

Answer 9

I'm sorry how do i get my 25.5 g into moles?

Answer 10

25.5g x 1mol / 180g = 0.1417 mol

Answer 11

\(\sf moles=\dfrac{mass}{Molar~mass}\)

yup thats it

Answer 12

m= 0.1417mol/ 0.398kg= 0.3359m

Answer 13

0.3559m x -1.86 C/m= 0.662C ??

Answer 14

the constant has a negative sign, so the answer should be negative

Answer 15

is that right?

Answer 16

you need a negative sign though

Answer 17

otherwise it's good

Answer 18

-0.662 C

Answer 19

yeah thats it

Answer 20

Thanks!

Answer 21

np!